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Can someone show me how to write complex numbers in standard form? I missed a few days of class and do not have the text book. Answering a simple question like the one below would help

Write the complex number in standard form. $6 + \sqrt{−16}$

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I hope that you get the textbook soon. –  Jonas Meyer May 29 '12 at 3:13
    
possible duplicate of Complex Numbers in Standard Form and other assorted problems –  Gerry Myerson May 29 '12 at 3:23
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Have you tried talking to your teacher, or your fellow students? Those should be your first recourses when making up missed classes. –  Chris Eagle May 29 '12 at 9:53
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2 Answers 2

$\sqrt{-1}$ is written as just $i$ for "imaginary".

$\sqrt{-x}$ can be factored as $\sqrt{-1\vphantom{x}}\sqrt{x} = i\sqrt{x}$.

"Standard form" for complex numbers is $a + bi$ where $a$ and $b$ are real numbers. If $a$ or $b$ is 0, you omit that part. For example, you write $3 + 0i$ as just $3$, and $0 + 3i$ as just $3i$.

For your example, you have $6+\sqrt{-16} = 6 + i\sqrt{16} = 6 + 4i$. The "standard form" is $6+4i$.

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You have to be carefull, when defining the imaginary unit as you did, see en.wikipedia.org/wiki/Imaginary_unit#Proper_use - the correct way is, to say that (i) is defined by (i^2 = -1), which is note the same as your definition (to be precise, the root is not defined for complex numbers). –  Ronny Dec 16 '13 at 14:21
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Note that $i^2=-1$, so $\sqrt{-16}=\sqrt{i^216}=i4. $ Hence $6+\sqrt {-16}=6+i4{}$

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Who voted this down? Can you explain why? –  smanoos May 29 '12 at 3:14
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@Mark: $a+ib$ and $a+bi$ are both explicitly mentioned as standard form in some textbooks, and regardless I don't think that a teacher would be reasonable to not accept $6+i4$ as standard form. –  Jonas Meyer May 29 '12 at 3:19
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$a+ib$ is fine. $6+i4$ is bizarre. –  Gerry Myerson May 29 '12 at 3:24
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@GerryMyerson. It is not bizarre provided you know what it means. The reason why some textbooks prefer this style is that in case of squre roots the meaning can be misleading. For example $i\sqrt{3}$ might be preferred over $\sqrt{3}i$. This is because the latter appears as if one is finding the square root of the product of $3$ and $i$. –  smanoos May 29 '12 at 3:34
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$i\sqrt3$ is fine. $i4$ is bizarre. It looks like someone meant $i_4$ but couldn't find the underscore key. –  Gerry Myerson May 29 '12 at 3:41
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